Ptolemy's theorem

What the hell you talkin' about?

Quite simple, really. If you have a cyclicquadrilateral (a four sided figure with its corners on a circle), then the sum of the products of the lengths of opposite sides equals the product of the length of the diagonals. The converse is also true. If a quadrilateral obeys the equation, then it's cyclic.

Sounds like you're making this up...

No, really. It's quite simple to prove. Watch. Say we add a point called M on the diagonal BD in such a place that the angle ACB = MCD.

We also have that the angle CAB = CDB as they lie on the same arc. Now we have two similar triangles, ABC and DMC. This means that |MD|/|DC| = |AB|/|AC| or |AB|.|CD| = |AC|.|MD| . By considering the triangles BMC and ADC in an identical way we get |AD|.|BC| = |AC|.|MB| . Adding the two equations gives us our original statement: |AB|.|CD| + |AD|.|BC| = |AC|.|MD| + |AC|.|MB| = |AC|(|MD| + |MB|) = |AC|.|BD| . Which is nice.